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What is the greatest value of n such that 30!6n is an integer?

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What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 29 Apr 2016, 16:07
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Re: What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 29 Apr 2016, 21:04
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Bunuel wrote:
What is the greatest value of n such that 30!/6^n is an integer?

A. 11
B. 12
C. 13
D. 14
E. 15


This Q basically asks us the number of 6s in 30!..
and to find 6s in 30!, we require to find the biggest prime number of 6 in 30!..

so number of 3s in 30! = \([\frac{30}{3}]+[\frac{30}{3^2}]+[\frac{30}{3^3}] = 10+3+1 = 14\)
ans D
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Re: What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 25 Jul 2016, 06:08
Bunuel wrote:
What is the greatest value of n such that 30!/6^n is an integer?

A. 11
B. 12
C. 13
D. 14
E. 15


\(6^n\) can be written as \((2*3)^n\) = \(2^n *3^n\)

Now if we can figure how many 2 and 3 are present in 30 ! then we can have them removed. Cancelling(dividing) n number of 2 and 3 will remove n number of 6 (because 6 is a multiple of 2 and 3)
Therefore removing \(2^n\)and\(3^n\)will remove \(6^n\) and thus we can get the highest number of n that can be removed from 30!

2's in 30! =\(2^{26}\)
3's in 30!= \(3^{14}\)
so we can easily remove \(2^{14} *3^{14}\) from 30!
so \(6^{14}\) can be removed
therefore n=14
ANSWER IS D
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What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 25 Jul 2016, 07:02
chetan2u wrote:
Bunuel wrote:
What is the greatest value of n such that 30!/6^n is an integer?

A. 11
B. 12
C. 13
D. 14
E. 15


This Q basically asks us the number of 6s in 30!..
and to find 6s in 30!, we require to find the biggest prime number of 6 in 30!..

so number of 3s in 30! = \([\frac{30}{3}]+[\frac{30}{3^2}]+[\frac{30}{3^3}] = 10+3+1 = 14\)
ans D


I knew that we could use \([\frac{n}{5^1}]+[\frac{n}{5^2}]+[\frac{n}{5^3}] ...\) for trailing of zeros of n!

But is it true for counting any prime number of n! ?
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Re: What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 25 Jul 2016, 08:51
1
matsuda wrote:
chetan2u wrote:
Bunuel wrote:
What is the greatest value of n such that 30!/6^n is an integer?

A. 11
B. 12
C. 13
D. 14
E. 15


This Q basically asks us the number of 6s in 30!..
and to find 6s in 30!, we require to find the biggest prime number of 6 in 30!..

so number of 3s in 30! = \([\frac{30}{3}]+[\frac{30}{3^2}]+[\frac{30}{3^3}] = 10+3+1 = 14\)
ans D


I knew that we could use \([\frac{n}{5^1}]+[\frac{n}{5^2}]+[\frac{n}{5^3}] ...\) for trailing of zeros of n!

But is it true for counting any prime number of n! ?


Yes, you can find the power of a prime in factorial this way. Check the links below.



Check Trailing Zeros Questions and Power of a number in a factorial questions in our Special Questions Directory.

Hope it helps.
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Re: What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 28 Jul 2016, 11:38
Here is what i did=>
t get the number of 6 we need to count the number of 3's as the number of 2 will always be adequate.
=> 3*1
=> 3*2
=> 3^2*1
=>3*4
=>3*5
=>3^2*2
=>3*7
=>3*8
=>3^3
=>3*10

Number of three's=> 14
Smash that D
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Re: What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 29 Jul 2016, 09:30
30! needs to be divisible by 2 and 3 simultaneously

How many 2's are there ? so that the division leads to be an integer

30/2 + 30/(2^2)+30/(2^3)+30(2^4) = 15+7+3+1=26

How many 3's are there ? so that the division leads to be an integer

30/3 + 30/(3^2)+30/(3^3)= 10+3+1=14

How many 3's and 2's are there so that the division leads to be an integer

Minimum or common between 26 and 14. Answer is 14.

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Re: What is the greatest value of n such that 30!6n is an integer?  [#permalink]

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New post 24 Aug 2018, 06:36
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Re: What is the greatest value of n such that 30!6n is an integer? &nbs [#permalink] 24 Aug 2018, 06:36
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